Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:

import java.util.*;

public class Main{
	
	public static double find(double arr[],double dp[][],int tail,int i,int n)
	{
		if(i>=n)
			return 1;
		
		if(dp[tail][i]==-1)
		{
			double op1=(arr[i])*find(arr,dp,tail,i+1,n);
			double op2=0;
			if(tail<n/2)
				op2=(1-arr[i])*find(arr,dp,tail+1,i+1,n);
			
				dp[tail][i]=op2+op1;
				
		}
		
			return dp[tail][i];
	}
    public static void main(String[] args) 
    {
       Scanner s=new Scanner(System.in);
       int n=s.nextInt();
       double arr[]=new double[n];
       for(int i=0;i<n;i++)
    	   arr[i]=s.nextDouble();
       
       double dp[][]=new double[n/2+1][n];
       for(int i=0;i<=n/2;i++)
       {
    	   for(int j=0;j<n;j++)
    		   dp[i][j]=-1;
       }
       System.out.print(find(arr,dp,0,0,n));
    }
}